Question

The probability a student will drop an online course is 40%. In a class of 20, what is the probability that more than 5 students will drop the course?

Answer 1

The situation in the above question describes a binomial distribution.

In probability theory and statistics, the binomial distribution with parameters n (sample size) and p (probability of success) is the discrete probability distribution of the number of successes in a sequence of n independent experiments having two possible outcomes, for instance asking a yes–no question.

The binomial probability of an event (X) occuring is given by

`P(X)=^nC_x(p)^x(q)^(n-x)`
where n is the sample size, p is the predetemined success probability and q is the failure probability given by q = 1 - p.

Given that the probability a student will drop an online course is 40%. This means that our probability of success is 40% or 0.4 and the probability of failure is 1 - 0.4 = 0.6

Given that the size of the class is 20, this means that n = 20.

The probability that more than 5 students will drop the course given by

`P(X \ \textgreater \ 5) = 1 - P(X \leq 5) \\ \\ =1-[P(0)+P(1)+P(2)+P(3)+P(4)+P(5)] \\ \\ P(0)=^(20)C_0(0.4)^0(0.6)^(20)=1(1)(0.0000365616)=0.0000365616 \\ \\ P(1)=^(20)C_1(0.4)^1(0.6)^(19)=20(0.4)(0.0000609360)=0.000487488 \\ \\ P(2)=^(20)C_2(0.4)^2(0.6)^(18)=190(0.16)(0.000101560)=0.00308742 \\ \\ P(3)=^(20)C_3(0.4)^3(0.6)^(17)=1,140(0.064)(0.000169267)=0.012350 \\ \\ P(4)=^(20)C_4(0.4)^4(0.6)^(16)=4,845(0.0256)(0.000282111)=0.034991`

`P(5)=^(20)C_5(0.4)^5(0.6)^(15)=15,504(0.0124)(0.000470185)=0.090393`

Thus,

`P(X \ \textgreater \ 5) =1-[0.0000365616+0.000487488+0.00308742 \\ +0.012350+0.034991+0.090393] \\ \\ =1-0.141345 \\ \\ =0.858655 \\ \\ \approx86\%`

Therefore, the probability that more than 5 students will drop the course is about 86%.